Tool to generate permutations of items, the arrangement of distinct items in all possible orders: 123,132,213,231,312,321.
Permutations - dCode
Tag(s) : Combinatorics
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Permutations
Permutations Generator
Permutations/Anagrams Calculator
Answers to Questions (FAQ)
What is a permutation? (Definition)
In Mathematics, item permutations consist in the list of all possible arrangements (and ordering) of these elements in any order.
Example: The three letters A,B,C can be shuffled (anagrams) in 6 ways: A,B,C B,A,C C,A,B A,C,B B,C,A C,B,A
Permutations should not be confused with combinations (for which the order has no influence) or with arrangements also called partial permutations (k-permutations of some elements).
How to generate permutations?
The best-known method is the Heap algorithm (method used by this dCode's calculator).
Here is a pseudo code source : function permute(data, n) {
if (n = 1) print data
else {
for (i = 0 .. n-2) {
permute(data, n-1)
if (n % 2) swap(data[0], data[n-1])
else swap(data[i], data[n-1])
permute(data, n-1)
}
}
}
Permutations can thus be represented as a tree of permutations: 
Another algorithm, named JohnsonâTrotter, generates all permutations by changing only one pair of adjacent elements at each step.
How to count permutations?
Counting the total number of permutations of $ n $ distinct elements is given by the factorial: $$ n! = n \times (nâ1) \times (nâ2) \times \cdots \times 2 \times 1 $$
Example: For $ 4 $ elements, the number of combinations is $ 4! = 24 $
How to count distinguishable permutations?
When a set contains repeated elements, some permutations are identical. The number of distinct permutations is then given by the formula: $$ \frac{ n! } { n_1! \times n_2! \times \cdots \times n_i! } $$ with $ n_i $ the number of occurrences of each repeated element.
In other words, having a repeated item involves a division of the number of permutations by the number of permutations of these repeated items.
Example: DCODE 5 letters have $ 5! = 120 $ permutations but contain the letter D twice (these $ 2 $ letters D have $ 2! $ permutations), so divide the total number of permutations $ 5! $ by $ 2! $: $ 5!/2!=60 $ distinct permutations.
Source code
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